Answer:
The new volume is 480.75L.
Explanation:
The given information from the exercise is:
- Initial amount of moles (n1): 5.47 moles
- Initial volume (V1): 186.5L
- Final amount of moles (n2): 14.1 moles
1st) We have to replace the values of V1 and n1 in the Idea Gases formula:
[tex]\begin{gathered} P*V_1=n_1*R*T \\ P*186.5L=5.47moles*0.082\frac{atm*L}{mol*K}*T \\ \frac{186.5L}{5.47moles*0.082\frac{atm*L}{mol*K}}=\frac{T}{P} \\ 415.8\frac{K}{atm}=\frac{T}{P} \end{gathered}[/tex]Now we have a value for the relation T/P that remains constant.
2nd) Finally, we have to replace the value of the relation T/P and the amoun of moles n2:
[tex]\begin{gathered} P*V=n_2*R*T \\ P*V=14.1moles*0.082\frac{atm*L}{mol*K}*T \\ V=14.1moles*0.082\frac{atm*L}{mol*K}*\frac{T}{P} \\ V=14.1moles*0.082\frac{atmL}{molK}*415.8\frac{K}{atm} \\ V=480.75L \end{gathered}[/tex]So, the new volume is 480.75L (rounded to 481L).
